Sheldon Ross suggests that for Poisson random variables $X$ and $Y$ with parameters $\lambda$ and $\mu$ respectively, $P(X < Y) = \lambda/(\lambda + \mu)$, which is the same for the probability of two Exponential random variables with the same respective intensity parameters. Not sure how he justifies it, although for a Poisson PROCESS, it makes sense. Any thoughts on why this may be?
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user39646Sep 6 '13 at 21:12

4 Answers
4

To get an estimate one simply finds that the dominating element among
$$A_k= \mathbb{P}(Y=k+1)\mathbb{P}(X=k) = e^{-1-\lambda} \frac{\lambda^k}{k!(k+1)!}$$
is $A_{\sqrt{\lambda}}$ which gives roughly $e^{2\sqrt{\lambda}-\lambda}$. Probably there's also a Poly($\lambda$) factor here.

Oh, and $F'(\lambda)$ is simply $\mathbb{P}(X=Y)$ since the probability of adding 1 when increasing the intensity of a Poisson RV by $\epsilon$ is $\epsilon$. In the special case $\lambda=1$ we get
$$F'(1)=e^{-2} \sum_{k=0}^\infty \frac{1}{(k!)^2}$$

Numerically, $\lim_{\lambda \to 0} F^\prime(\lambda) = 1/e$. Heuristically this should be true because to have $X > Y$ when $\lambda$ is very small, the most likely case will be $X = 1, Y = 0$ by far; that occurs with probability $\lambda e^{-\lambda} e^{-1}$.

Maple gives an explicit formula for $F^\prime(1)$ involving sums; evaluating gives $F^\prime(1) = 0.3085083225$, and the Inverse Symbolic Calculator says this is somehow to $I_0(2) e^{-2}$. I'm not sure how to prove this at all but maybe knowing the answer helps?

By simple computations : The definition of the modified Bessel function of the first kind yields
$$
I_k(\lambda)=\sum_{n\geq 0}\frac{1}{n!(n+k)!}\left(\frac{\lambda}{2}\right)^{2n+k}
$$
so that we get (the sums transpositions are clearly allowed)
$$F(\lambda)=e^{-\lambda-1}\sum_{k\geq 1}\sum_{n\geq 0}\frac{\lambda^{k+n}}{n!(k+n)!}=e^{-\lambda-1}\sum_{n\geq 1}a_n\lambda^n \qquad \mbox{where}\qquad a_n=\frac{1}{n!}\sum_{k=0}^{n-1}\frac{1}{k!}.$$ Thus, deriving under the sign sum